Question

Find the ratio in which the \(x\)-axis divides the line segment joining the points \((6, 5)\) and \((-4, -1)\). Also find the point of intersection.

Hint:

A quick formula for the ratio in which \(x\)-axis divides \((x_1, y_1)\) and \((x_2, y_2)\) is \(-y_1 : y_2\).
Here, \(-5 : -1\) which simplifies to \(5 : 1\).

Answer 1

We are asked to find the ratio in which the \(x\)-axis divides the line segment joining the points \(A(6, 5)\) and \(B(-4, -1)\), and also the coordinates of the point of intersection.
Step 1: Let the required point be P(x, 0)
- Let the \(x\)-axis divide AB in the ratio \(k:1\), where k is the ratio AP : PB.
- Using the section formula, the coordinates of P dividing AB internally in the ratio k:1 are given by:
\[ P = \left( \frac{k x_2 + x_1}{k+1}, \frac{k y_2 + y_1}{k+1} \right) \]
Here, \(x_1 = 6, y_1 = 5, x_2 = -4, y_2 = -1\).
Step 2: Use the y-coordinate to find k
- Since P lies on the x-axis, its y-coordinate = 0:
\[ \frac{k(-1) + 5}{k+1} = 0 \]
\[ \frac{-k + 5}{k+1} = 0 \]
\[ -k + 5 = 0 \implies k = 5 \]
Step 3: Find the coordinates of P
- Using the x-coordinate formula:
\[ x = \frac{k x_2 + x_1}{k+1} = \frac{5(-4) + 6}{5+1} = \frac{-20 + 6}{6} = \frac{-14}{6} = -\frac{7}{3} \]
- So, P = \(\left(-\frac{7}{3}, 0\right)\)
Step 4: Write the ratio
- The x-axis divides the line segment AB in the ratio AP : PB = 5 : 1
Answer:
- Ratio = 5 : 1
- Point of intersection = \(\left(-\frac{7}{3}, 0\right)\)